c) \( \lim _{x \rightarrow+\infty} \frac{4 x+3}{\sqrt{2 x^{2}+3 x}} \)

To evaluate the limit \( \lim _{x \rightarrow+\infty} \frac{4 x+3}{\sqrt{2 x^{2}+3 x}} \), we can start by simplifying the expression. First, we observe that as \( x \) approaches infinity, the highest degree terms in the numerator and denominator dominate the behavior of the function. Thus, we can factor out \( x \) from both the numerator and denominator: \[ \lim _{x \rightarrow+\infty} \frac{4 x + 3}{\sqrt{2 x^2 + 3 x}} = \lim _{x \rightarrow+\infty} \frac{x(4 + \frac{3}{x})}{\sqrt{x^2(2 + \frac{3}{x})}} \] Now simplifying further, we have: \[ = \lim _{x \rightarrow+\infty} \frac{x(4 + \frac{3}{x})}{x\sqrt{2 + \frac{3}{x}}} = \lim _{x \rightarrow+\infty} \frac{4 + \frac{3}{x}}{\sqrt{2 + \frac{3}{x}}} \] As \( x \) approaches infinity, \( \frac{3}{x} \) approaches \( 0 \): \[ = \frac{4 + 0}{\sqrt{2 + 0}} = \frac{4}{\sqrt{2}} = \frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \] Thus, the final answer is: \[ \lim _{x \rightarrow+\infty} \frac{4 x + 3}{\sqrt{2 x^{2}+3 x}} = 2\sqrt{2} \]